Question

A particle is moving with a uniform acceleration its position time t seconds is given in metres by the relation x =3+4t+5t^2.Calculate the magnitude of its position at t =5 second ​

Answer 1

Answer:

Explanation:

v = dx/dt

v = d(2+5t+7t²)/dt

v = 5 + 14t

i) initial velocity  

= velocity at t = 0

= 5m/s

ii) t = 4

v = 5 + 56

= 61m/s

iii) a = dv/dt  

= 14 m/s²

iv) x at t=5

x = 2 + 5×5 + 7×5²

= 2 + 25 + 175

= 202 m

Answer 2

Here is ur answer:

At t= 5s, position is given as,

$x = 3 + 4(5) + 5( {5)}^{2}$

= 3+ 20+ 5× 25

=3+ 20+ 125

=148m